Agenda

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Angular Momentum of Point Mass

• Assume a point mass \(m\) that has a momentum \(\mv{p}^o\)
• Assume a vector from the origin of the observer's frame \(O\) to the point mass \(\mv{r}^o\)
• Angular momentum: \[ \mv{L}^o=\mv{r}^o\times \mv{p}^o \]

Rotational Inertia Preparation

\(\mv{v}\) can be decomposed into tangential velocity \(\mv{v}_t\) and radial velocity \(\mv{v}_r\)

\(\mv{r}\times \mv{v}=\mv{r}\times (\mv{v}_t+\mv{v}_r)=\mv{r}\times \mv{v}_t=\mv{r}\times (\mv{\omega} \times \mv{r})\)

Rotational Inertia of Point Mass

\[ \begin{aligned} \mv{L}^o&=\mv{r}^o\times \mv{p}^o=\mv{r}^o \times (m\mv{v}^o)=m \mv{r}^o \times (\mv{\omega}^o \times \mv{r}^o)\\ &=-m \mv{r}^o\times (\mv{r}^o\times \mv{\omega}^o)=-m[\mv{r}^o][\mv{r}^o]\mv{\omega}^o \end{aligned} \]
Angular momentum depends on the choice of the observer's frame!
• Recall that a momentum, such as \(\mv{p}\), is a product of inertia and velocity
• We define the rotational inertia similarly. The rotation inertia for a point mass is \[ \mv{I}^o=-m[\mv{r}^o][\mv{r}^o] =\begin{bmatrix} m(r_y^2+r_z^2) & -mr_xr_y & -mr_xr_z\\ -mr_xr_y & m(r^2_x+r^2_z) & -mr_yr_z\\ -mr_xr_z & -mr_yr_z & m(r^2_x+r^2_y) \end{bmatrix} \in\mathbb{R}^{3\times 3} \]
• Then, \[\mv{L}^o=\mv{I}^o\mv{\omega}^o \]

Angular Momentum and Inertia of Rigid Body

• Let us view rigid body as a system of particles whose relative positions are fixed (no deformation).
• Define the angular momentum of a body by aggregating from volume elements: \[ \mv{L}^o=\int_{x^o \in B} \d{\{\mv{r}^o(x) \times \mv{p}^o(x^o)\}}=\int_{x^o\in B} \d{\{\mv{r}^o(x) \times m(x^o) \mv{v}^o(x^o)\}} \]
• One more step: \[ \mv{L}^o=\int_{x^o\in B} -\d{\{m^o(x^o) [\mv{r}^o(x^o)][\mv{r}^o(x^o)]\mv{\omega}^o\}}=\left(\int_{x^o\in B} -\d{\{m(x^o) [\mv{r}^o(x^o)][\mv{r}^o(x^o)]\}}\right)\mv{\omega}^o \]

Angular Momentum and Inertia of Rigid Body

• Particularly, if we choose the origin of the observer's frame \(O\) at the center of mass: \[ \mv{L}^{b}=\mv{I}^{b}\mv{\omega}^b \tag{body angular momentum} \] where \[ \mv{I}^{b}=\int_{x^b\in B} -\d{V}\{\rho(x^b) [\mv{r}^b(x^b)][\mv{r}^b(x^b)]\}\tag{body inertia} \] and center of mass \[ x^o_{cm}=\frac{\int \mv{r}^o \rho d\mathbf{V}}{\int\rho d\mathbf{V}} \tag{center of mass} \]
• Since \(\cal{F}_{b(t)}\) is tightly binded to the body, \(\mv{I}^b\) does not change w.r.t. time and is a basic property of the object.

Computation of Rigid Body Inertia

\[ \begin{aligned} \mv{I}^b&=\int_{x^b\in B} -\d{V}\rho(\mv{x}^b) [\mv{r}^b(\mv{x}^b)][\mv{r}^b(\mv{x}^b)]\\ &=\begin{bmatrix} \int\rho(r_y^2+r_z^2)d\mv{V} & -\int\rho r_xr_yd\mv{V} & -\int\rho r_xr_zd\mv{V}\\ -\int\rho r_xr_yd\mv{V} & \int\rho (r_x^2+r_z^2)d\mv{V} & -\int\rho r_yr_zd\mv{V}\\ -\int\rho r_xr_zd\mv{V} & -\int\rho r_yr_zd\mv{V} & \int\rho(r_y^2+r_x^2)d\mv{V}\\ \end{bmatrix} \end{aligned} \]
• Given uniform density, the integral can be computed analytically for watertight meshes

Fast Inertia Computation

• Divergence theorem! Let \(\mv{F}:\mathbb{R}^3\to\mathbb{R}^3\), \(\int_V\nabla\cdot \mv{F} dV = \oint_S \mv{F}\cdot \mv{n} d S\)
• An example: a term of \(\mv{I}\), which is \(-\rho \int_\mv{V}r_yr_zd\mv{V}\)
  Let \(\mv{F}(r_x,r_y,r_z) = \begin{bmatrix}r_xr_yr_z &0 & 0\end{bmatrix}^T\) \[\nabla\cdot\mv{F} = r_yr_z\] The integral becomes \[\oint_S \mv{F}\cdot\mv{n} dS\] Now we only need to do 2D integral over triangles.

Mass Properties

• Observe \( \mv{I}^b=\int_{\mv{r}^b\in B} -\d{\mv{V}}\rho(\mv{r}^b)[\mv{r}^b][\mv{r}^b] \)
• Although the origin is always at the center of mass, if we change the orientation of body frame axes, \(\mv{I}^b\) may change!
• How will it change, then?

• If we rotate the frame by \(R^T\) and obtain a new frame \(b'\), then \[ \begin{aligned} {\mv{I}^b}'&=\int_{\mv{r}^b\in B} -\d{\mv{V}}\rho(\mv{r}^b)[R\mv{r}^b][R\mv{r}^b]=\int_{\mv{r}^b\in B} -\d{\mv{V}}\rho(\mv{r}^b)R[\mv{r}^b][\mv{r}^b]R^T =R \mv{I}^b R^T \end{aligned} \] where the second equality follows \([Rr]=R[r]R^T\) for \(R\in\mathbb{SO}^3\). Again, similarity transformation!

Conclusion: Rigid-transformation does not change the eigen properties of \(\mv{I}^b\)

Mass Properties

Quiz

Suppose an object is moving in space (rotating and translating), which of the following quantities may change during the motion. (Assume all quantities are measured w.r.t. a static spatial frame)

• A. principal axes (observed from the spatial frame)
• B. \(x_{cm}\) (observed from the spatial frame)
• C. \(m\)
• D. \(I_1,I_2,I_3\)

Torque

Torque

Euler Equation

\[ \begin{align*} % \mv{f} &= \frac{\d{\mv{p}}}{\d{t}}=\frac{\d{(m\mv{v})}}{\d{t}}=m\mv{a} \tag{linear motion}\\ \mv{\tau}^b &= \frac{\d{\mv{L}^b}}{\d{t}}=\frac{\d{(\mv{I}^b\mv{\omega}^b)}}{\d{t}}=\frac{\d{\mv{I}^b}}{\d{t}}\mv{\omega}^b+\mv{I}^b\frac{\d{\mv{\omega}^b}}{\d{t}}=\mv{\omega}^b\times \mv{I}^b\mv{\omega}^b + \mv{I}^b\dot{\mv{\omega}^b} \end{align*} \]

• We used \(\dot{\mv{I}^b}=[\mv{\omega}]\mv{I}^b\) without proof
• An observation (example): Even if there is no torque input, if the object has a non-zero angular velocity \(\mv{\omega}^b\), then it may still have an angular acceleration \(\dot{\mv{\omega}^b}\)
  • When \(\mv{\omega}^b \not\parallel \mv{I}^b\mv{\omega}^b\), i.e., \(\mv{\omega}^b\) is not along the eigenvector of \(\mv{I}^b\), \(\mv{\omega}^b\) will NOT keep unchanged
  • \(\mv{\omega}^b\) will not converge (\(\dot{\omega}^b\) will never be zero). Its trajectory will form a periodic curve
  • \(\mv{L}^b=\mv{I}^b\mv{\omega}^b\) is conserved

Euler Equation

\[ \mv{\tau}^b=\mv{\omega}^b\times \mv{I}^b\mv{\omega}^b + \mv{I}^b\dot{\mv{\omega}^b} \tag{angular motion} \]

A numerical experiment in SAPIEN for \(\mv{\tau}^b=0\)

(this is illustrative and there are numerical errors)